Thursday, January 4, 2007

2007. 4 and 39

Problem (4)
Difficulty: first page.

Problem 39
Note: Remainder.
Difficulty: Hard. (last page)

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest
Yeah, I know, it should read "larger" and "smaller" but I didn't have the energy to fix it.

answer 4 and answer 39.


  1. I thought of a way I can do this but I'm honestly not 100% sure in this solution:

    1st factor out x^3 from both p(x) and q(x). Then, since we are dividing, divide through by x^3. The fraction should be:

    p(x)/q(x)= (x^2007+x^1998+x^1989+...+x^18+x^9+1)/(x^3-1).

    Now, create a dummy variable u=x^3 (this is where my uncertainty comes in-- think this should work but unsure). So

    p(u)/q(u)= (u^669+u^663+u^660+...+u^6+u^3+1)/(u-1).

    It was nice of them to tell us how many terms p(u) has (224) because we can now use the remainder theorem and say p(1) will equal the remainder. p(1)=224(1)=224.

    This leads to the remainder of
    224/(u-1) or 224/(x^3-1)

  2. The answer still doesn't make sense to me. If the remainder was 224x^3 and I divide that by long division by x^3-1 doesn't it have a remainder of 224?

    (x^3-1)| 224x^3

  3. Good thing you insisted. I messed that up. Must revisit. I put a better long division in the answer sheet, right after the wrong method I had there earlier (which matched the right answer by the "two wrongs DO make a right" method).

    Essentially, it boils down to when you are finished, you will have 224/u-1 but we can't leave it there. We removed a x³ from numerator and denominator so the last bit is really 224x³ / (x^6-x^3) so the remainder is really 224x³.

    Good thing I wasn't taking the test for real.